If $\Bbb{P}$ and $\Bbb{Q}$ are two probability measures such that $\Bbb{Q} \ll \Bbb{P}$ (i.e $\Bbb{Q}$ is absolutely continuous with respect to $\Bbb{P}$), can something be said about the expectation of a random variable $X$ with respect to $\Bbb{P}$ and the expectation of that same random variable with respect to $\Bbb{Q}$ ? For instance, would we have $\Bbb{E}^\Bbb{Q}[X] \leq \Bbb{E}^\Bbb{P}[X]$ ?
From Radòn-Nikodym theorem we would have that $\Bbb{E}^\Bbb{Q}[X] = \Bbb{E}^\Bbb{P}[\frac{d\Bbb{Q}}{d\Bbb{P}}X]$ but I am not sure one can say something about the value of $\frac{d\Bbb{Q}}{d\Bbb{P}}$.
Thank you for your help.
You really should use $\mathbf{Q}$ or simply $Q$ instead of $\Bbb{Q}$ which usually means the rational numbers.
That being said , no you cannot really say much and also you cannot really state any trichotomic result like that.
For example, consider $P$ to be a Gaussian $N(\mu_{1},\sigma^{2})$ Distribution on $\Bbb{R}$ and $Q$ to be an $\text{Exponential}(\mu_{2})$ distribution on $\Bbb{R}$ . Fix a common sigma algebra on $\Bbb{R}$ namely the Borel Sigma algebra. Then $Q<<P$ simply because $P(A)=0\iff\lambda(A)=0$ which always means $Q(A)=0$ for any Borel set. ($\lambda$ is the Lebesgue Measure)
Now consider any Borel measurable function say $f:(\Bbb{R},\mathcal{B}(\Bbb{R}))\to\Bbb{R}$ . For simplicity consider $f(x)=x$ . Then $f$ is a real valued random variable from both the probability spaces $(\Bbb{R},\mathcal{B}(\Bbb{R}),P)$ and $(\Bbb{R},\mathcal{B}(\Bbb{R}),Q)$ and now you consider it's expectation with respect to sticking by your notation
$E^{Q}(f(x))=\int_{\Bbb{R}}x\,dQ(x)=\mu_{2}$ and you also have $E^{P}(f(x))=\int_{\Bbb{R}}x\,dP(x)=\mu_{1}$ .
Now notice that I have chosen $\mu_{1}$ and $\mu_{2}$ arbitrarily and so you can easily see that there need be no relation between them.